By Simpson's rule, the value of ∫12dxx dividing the inte

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 Multiple Choice QuestionsMultiple Choice Questions

351.

cos3x . elogsinxdx is equal to

  • - sin4x4 + c

  • - cos4x4 + c

  • esinx4 +c

  • None of the above


352.

The value of 0π2cos3x + 12cosx - 1dx is

  • 2

  • 1

  • 12

  • 0


353.

The value of 01tan-12x - 11 + x - x2dx is

  • 1

  • 0

  • - 1

  • None of the above


354.

By the application of Simpson's one - third rule for numerical integration, with two subintervals, the value of 01dx1 +x is

  • 1736

  • 1725

  • 2536

  • 1724


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355.

xe - 1 + ex - 1xe + exdx is equal to

  • logxe + ex +c

  • elogxe + ex +c

  • 1elogxe + ex +c

  • None of the above


356.

The value of π4π2exlogsinx + cotxdx is

  • eπ4log2

  • - eπ4log2

  • 12eπ4log2

  • - 12eπ4log2


357.

Considering four sub-intervals, the value of 0111 +xdx by Trapezoidal rule, is

  • 0.6870

  • 0.6677

  • 0.6977

  • 0.5970


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358.

By Simpson's rule, the value of 12dxx dividing the interval (1, 2) into four equal parts, is

  • 0.6932

  • 0.6753

  • 0.6692

  • 7.1324


A.

0.6932

h = 2 - 14 = 14Now, x0 = 1, x1 = 1 +14, x2 = 1 + 2 × 14,         x3 = 1 + 3 × 14, x4 = 1 + 4 × 14i.e.,   x0 = 1, x1 = 1.25, x2 = 1.5, x3 = 1.75, x4 = 2     y0 = 1, y1 = 0.8, y2 = 0.667, y3 = 0.571, y4 = 0.5 Using Simpson's 13rd rule12dxx = 1121 + 0.5 + 40.8 + 0.571 + 20.667           = 1121.5 + 5.484 + 1.334           = 1128.318 = 0.6932


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359.

The value of xsinxsec3xdx is

  • 12sec2x - tanx + c

  • 12xsec2x - tanx + c

  • 12xsec2x + tanx + c

  • 12sec2x + tanx + c


360.

The value of 0xxsin3xdx is

  • 4π3

  • 2π3

  • 0

  • None of these


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